Question

If $(\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a; 0<x<1, a \neq 0$, then the value of $2x^2 - 1$ is :

• A. $\cos \left( \frac{2a}{\pi} \right)$
• B. $\sin \left( \frac{2a}{\pi} \right)$
• C. $\sin \left( \frac{4a}{\pi} \right)$
• D. $\cos \left( \frac{4a}{\pi} \right)$

Hint:

Recall trigonometric identities for inverse functions and double angles together. $2x^2 - 1$ is $\cos(2\cos^{-1} x)$ or $-\cos(2\sin^{-1} x)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Use the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$. The expression $2x^2 - 1$ is related to double angle formulas for cosine.
Step 2: Detailed Explanation:
Given: $(\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a$.
Using $A^2 - B^2 = (A - B)(A + B)$:
\[ (\sin^{-1} x - \cos^{-1} x)(\sin^{-1} x + \cos^{-1} x) = a \]
Since $\sin^{-1} x + \cos^{-1} x = \pi/2$:
\[ (\sin^{-1} x - \cos^{-1} x) \frac{\pi}{2} = a \implies \sin^{-1} x - \cos^{-1} x = \frac{2a}{\pi} \]
We have a system:
1. $\sin^{-1} x + \cos^{-1} x = \pi/2$
2. $\sin^{-1} x - \cos^{-1} x = 2a/\pi$
Adding them: $2\sin^{-1} x = \frac{\pi}{2} + \frac{2a}{\pi}$.
Let $\sin^{-1} x = \theta$, so $x = \sin \theta$.
We need $2x^2 - 1 = 2\sin^2 \theta - 1 = -\cos(2\theta)$.
From our result: $2\theta = \frac{\pi}{2} + \frac{2a}{\pi}$.
\[ -\cos(2\theta) = -\cos \left( \frac{\pi}{2} + \frac{2a}{\pi} \right) = - \left( -\sin \frac{2a}{\pi} \right) = \sin \left( \frac{2a}{\pi} \right) \]
Step 3: Final Answer:
The value of $2x^2 - 1$ is $\sin \left( \frac{2a}{\pi} \right)$.